AP CHEMISTRY
TOPIC 7: ACIDS & BASES, PART D
•
•
Polyprotic acids
Day 81:
•
Amphoteric substance
Acid-Base salts
1. Write out the stepwise dissociation reactions and the Ka expressions for the diprotic acid H2SO3 .
H2SO3 ↔ HSO3
HSO3
-1
2. Calculate the pH of a 0.10 M H2CO3 (
[H2CO3 ]
+
0.10 M
-x
0.10 M - x
I
C
E
-1
↔ SO3
-2
+ H
+1
+ H
+1
⎡⎣ HSO3−1 ⎤⎦ ⎡⎣ H +1 ⎤⎦
[ H 2 SO3 ]
⎡⎣ SO3−2 ⎤⎦ ⎡⎣ H +1 ⎤⎦
⎡⎣ HSO3−1 ⎤⎦
K a = 4.3 x 10 -7, K a = 4.3 x 10 -11 ).
1
2
[ H2O ]
↔
[ HCO3-1 ]
-
0
+x
x
= 4.3 × 10
−7
[ H3O +1 ]
0
+x
x
[ HCO ] [ H O ] x
x
=
=
=
[ H CO ]
0.10 − x 0.10
−1
K a1
+
+1
3
2
x 2 = ( 0.10 ) ( 4.3 × 10 −7 ) ,
2
3
x =
2
3
4.3 × 10−8 = 2.07 × 10 −4
[ H3O+1 ] = 2.1 x 10-4 M.
pH = - log ( 2.1 x 10-4 ) = 3.68
There is no need to do a second ICE chart since the pH will not change by significant amount… If you do
a second ICE chart, the concentration of the [ H+1 ] = 4.30 x 10-11 M. This concentration is so small it will
NOT affect the pH.
3. Given that the Ka value for acetic acid is 1.8 x 10-5 and the Ka value for hypochlorous acid is 3.5 x 10-8, which is the
stronger base, OCl-1 or C2H3O2-1? Explain why.
From the Ka values, acetic acid is a stronger acid than hypochlorous acid. Conversely, the conjugate base of acetic
acid, C2H3O2 -1, will be a weaker base than the conjugate base of hypochlorous acid, OCl –1, Thus, the
hypochlorite ion, OCl –1, is a stronger base than the acetate ion, C2H3O2 -1, In general, the stronger acid, the
weaker the conjugate base. This statement comes from the relationship Kw = Ka x Kb which holds true for all
conjugate acid-base pairs.
Also, you could calculate the Kb for each the of bases… A higher Kb equals the stronger base…
C2H3O2
-1
:
K w 1.00 × 10 −14
Kb =
=
= 5.56 × 10 −10
−5
Ka
1.8 × 10
;
OCl
–1
:
K w 1.00 × 10 −14
Kb =
=
= 2.86 × 10 −7
−8
Ka
3.5 × 10
4. Calculate the pH for 0.10 M CH3NH3Cl, Kb = 4.38 x 10-4 for CH3NH2.
CH3NH3Cl → CH3NH3+1 + Cl –1
( CH3NH3+1, conjugate acid of a weak base, Cl-1, conjugate base of a strong acid )
The chlorine ion, Cl-1, has no effect on the solution to make it basic or acidic.
K w 1.00 × 10 −14
=
= 2.28 × 10 −11
K b 4.38 × 10 − 4
Ka =
[ CH3NH3+1 ]
I
C
E
+
0.10 M
-x
0.10 M - x
[ H2O ]
U
[CH3NH2 ]
-
+
0
+x
x
0
+x
x
[ CH NH ] [ H O ] x
x
=
=
0
.
10
0
.10
x
−
[ CH NH ]
+1
K a = 2.28 × 10 −11 =
[ H3O +1 ]
3
2
2
3
2
+1
3
3
x 2 = ( 0.10 ) ( 2.28 × 10 −11 ) ,
x =
2.28 × 10 −12 = 1.51× 10−6
[ H3O +1 ] = 1.51 x 10-6 M
pH = - log ( 1.51 x 10-6 ) = 5.82
5. Calculate the pH for 0.050 M NaCN, Ka = 6.2 x 10-10 for HCN.
NaCN → Na +1 + CN –1
( Na +1, conjugate acid of a strong base, CN-1, conjugate base of a weak acid )
The sodium ion, Na +1, has no effect on the solution to make it basic or acidic.
Kb =
[ CN -1 ]
I
C
E
+
0.050 M
-x
0.050 M - x
K w 1.00 × 10 −14
=
= 1.61 × 10 −5
−10
Ka
6.2 × 10
[ H2O ]
U
[ HCN ]
-
0
+x
x
−5
2
2
−1
x 2 = ( 0.050 ) ( 1.61 × 10 −5 ) ,
x =
8.06 × 10−7 = 8.98 × 10 −4
[ OH -1 ] = 8.98 x 10-4 M
pOH = - log (8.98 x 10-4 ) = 3.05
pH = 14 - 3.05 = 10.95
[ OH -1 ]
0
+x
x
[ HCN ] [ OH ] x
x
=
=
=
[ CN ]
0.050 − x 0.050
−1
K b = 1.61 × 10
+
6. An unknown salt is either NaCN, NaC2H3O2, NaF, NaCl, NaNO2, or NaOCl. When 0.100 mol of the salt is dissolved
in 1.00 liter of water, the pH of the solution is 8.07. What is the identity of this salt? (use the chart below)
Answers:
All these salts contain Na +1 which has no acidic / basic properties and a conjugate base of a weak acid
( except for NaCl where Cl –1 is a neutral species. ) All conjugates bases of weak acids are weak bases
since Kb for these species are between Kw and one ( 1 ). To identify the species, we will use the data
given ( in the chart above ) to determine the Kb value for the weak conjugate base. From the Kb
value and the data above, we can identify the conjugate base present by calculating Ka value for the
weak acid.
[ A-1 ]
I
C
E
0.100 M
-x
0.100 M - x
+
U
[ H2O ]
-
Kb =
[ HA ]
+
0
+x
x
[
]
⎡⎣ OH −1 ⎤⎦
=
⎡⎣ A−1 ⎤⎦
HA
[ OH -1 ]
0
+x
x
x2
( 0.100 − x )
pH = 8.07 ( given from question )
pOH = 14.00 - 8.07 = 5.93
[ OH -1 ] = x = antilog ( - 5.93 ) = 1.17 x 10-6 M
( 1.17 ×10 )
−6
Kb =
2
( 0.100 − 1.17 ×10−6
)
= 1.38 × 10 −11 = For “Unknown
Salt”
Since we are NOT given Kb values for the Salts, we must find the Ka for the conjugate acid.
K w 1.00 × 10 −14
= 7.24 × 10 −4
=
Ka =
−11
K b 1.38 × 10
from the table above, this Ka value is closest to HF, Therefore the unknown salt is NaF.